Math Mastery — Algebra 2 & Geometry 2

A-01

Quadratic Functions Easy

Find the vertex of the parabola \( f(x) = 2x^2 - 8x + 5 \).
Enter answer as: (h, k)

⚡

Memory Key VERTEX FORMULA → h = −b/(2a) | then plug in for k

Worked Example — similar question

Given: \( f(x) = 3x^2 - 12x + 7 \)

→ \( h = \dfrac{-(-12)}{2 \cdot 3} = \dfrac{12}{6} = 2 \)

→ \( k = 3(2)^2 - 12(2) + 7 = 12 - 24 + 7 = -5 \)

✓ Vertex = (2, −5)

💡 Solution Steps

\( a=2, b=-8 \) → \( h = \dfrac{8}{4} = 2 \)
\( k = 2(4) - 8(2) + 5 = 8 - 16 + 5 = -3 \)
Vertex = (2, −3)

A-02

Complex Numbers Medium

Simplify: \( \dfrac{3+2i}{1-i} \) — write in \( a+bi \) form.
Enter: a+bi e.g. "2+3i"

⚡

Memory Key CONJUGATE TRICK → multiply top & bottom by CONJUGATE of denominator

Worked Example

\( \dfrac{2+i}{1-i} \cdot \dfrac{1+i}{1+i} \)

→ Numerator: \( (2+i)(1+i) = 2+2i+i+i^2 = 2+3i-1 = 1+3i \)

→ Denominator: \( 1^2+1^2 = 2 \)

✓ \( \dfrac{1+3i}{2} = \dfrac{1}{2} + \dfrac{3}{2}i \)

💡 Solution

Multiply by conjugate \(\dfrac{1+i}{1+i}\):
Numerator: \((3+2i)(1+i)=3+3i+2i+2i^2=3+5i-2=1+5i\)
Denominator: \(1+1=2\)
Answer: \(\dfrac{1}{2}+\dfrac{5}{2}i\)

A-03

Polynomial Division Medium

Use synthetic division to find the remainder when
\( p(x) = x^3 - 4x^2 + x + 6 \) is divided by \( (x - 2) \).
Enter a single integer.

⚡

Memory Key REMAINDER THEOREM → remainder = p(divisor root) → just plug in x=2!

Worked Example (Synthetic Division table)

Coefficients of \(x^3-4x^2+x+6\): 1 −4 1 6, divisor root = 2

↓   2   −4   −6

1   −2   −3   0 ← remainder

✓ Remainder = 0 (x−2 is a factor!)

💡 Remainder Theorem Shortcut

\( p(2) = 8 - 16 + 2 + 6 = 0 \)
Remainder = 0 → (x−2) is a factor of p(x)

A-04

Exponential Equations Medium

Solve for \( x \): \( 4^{x+1} = 8^{x-1} \)
Enter an integer.

⚡

Memory Key SAME BASE TRICK → convert both sides to power of 2, then set exponents equal

Method: Rewrite with base 2

\(4 = 2^2\) and \(8 = 2^3\)

→ \(2^{2(x+1)} = 2^{3(x-1)}\)

→ \(2x+2 = 3x-3\)

✓ \(x = 5\)

💡 Step-by-step

\(2^{2x+2}=2^{3x-3}\) → \(2x+2=3x-3\) → x=5

A-05

Logarithms Medium

Solve: \( \log_2(x+3) + \log_2(x-1) = 3 \)
Enter a single integer (check domain!).

⚡

Memory Key LOG PRODUCT RULE → log A + log B = log(A·B) → then CHECK DOMAIN (x > 1)

Steps

\(\log_2[(x+3)(x-1)] = 3\)

→ \((x+3)(x-1) = 2^3 = 8\)

→ \(x^2+2x-3=8\) → \(x^2+2x-11=0\)

→ \(x = \dfrac{-2\pm\sqrt{48}}{2}\) → check which is valid

💡 Solution

\(x^2+2x-11=0\) → \(x=\dfrac{-2\pm\sqrt{48}}{2}=-1\pm2\sqrt{3}\)
Check domain (\(x>1\)): only \(x = -1+2\sqrt{3} \approx 2.46\) is valid.
Answer: \(x = -1+2\sqrt{3}\)

A-06

Rational Functions Medium

What is the horizontal asymptote of
\( f(x) = \dfrac{3x^2 - 5}{x^2 + 2} \)?
Enter: y = [number]

⚡

Memory Key HA RULE: same degree top/bottom → y = (leading coeff top)/(leading coeff bottom)

3-Case Cheat Sheet

deg(top) < deg(bottom) → y = 0

deg(top) = deg(bottom) → y = ratio of leading coefficients

deg(top) > deg(bottom) → No H.A. (oblique instead)

💡 Solution

Both numerator and denominator have degree 2.
Leading coefficients: 3 (top) and 1 (bottom).
Horizontal Asymptote: y = 3

A-07

Sequences & Series Medium

The 3rd term of a geometric sequence is 12 and the 6th term is 96.
Find the common ratio \( r \).
Enter a single number.

⚡

Memory Key RATIO SHORTCUT → a₆/a₃ = r³ → solve for r by taking cube root

Steps

\(\dfrac{a_6}{a_3} = r^{6-3} = r^3\)

→ \(r^3 = \dfrac{96}{12} = 8\)

✓ \(r = \sqrt[3]{8} = 2\)

💡 Solution

\(r^3 = 96/12 = 8\) → r = 2

A-08

Systems of Equations Easy

Solve the system: \(\begin{cases} x^2 + y = 10 \\ x - y = 2 \end{cases}\)
How many solutions does this system have?
Enter: 0, 1, or 2

⚡

Memory Key NON-LINEAR SYSTEM → substitution method → solve quadratic → count valid solutions

Steps

From line 2: \(y = x-2\). Substitute into line 1:

→ \(x^2 + (x-2) = 10\) → \(x^2+x-12=0\)

→ \((x+4)(x-3)=0\) → \(x=-4\) or \(x=3\)

✓ Two valid solutions

💡 Solutions

\(x=3, y=1\) and \(x=-4, y=-6\) — 2 solutions

A-09

Radical Equations Medium

Solve: \( \sqrt{2x+1} = x - 1 \)
Check for extraneous roots! Enter a single integer.

⚡

Memory Key RADICAL EQUATION → square both sides → CHECK extraneous solutions (always!)

Steps

Square both sides: \(2x+1 = (x-1)^2 = x^2-2x+1\)

→ \(0 = x^2-4x\) → \(x(x-4)=0\) → \(x=0\) or \(x=4\)

→ Check \(x=0\): \(\sqrt{1}=1\neq 0-1=-1\) ✗ EXTRANEOUS

✓ \(x=4\): \(\sqrt{9}=3=4-1\) ✓

💡 Solution

x=0 is extraneous. Only x = 4 is valid.

A-10

Inverse Functions Hard

If \( f(x) = \dfrac{2x+3}{x-1} \), find \( f^{-1}(5) \).
Enter a single number (fraction OK).

⚡

Memory Key INVERSE → swap x and y, solve for y. Or: f⁻¹(5) means find x where f(x)=5

Shortcut Method: set f(x) = 5

\(\dfrac{2x+3}{x-1}=5\)

→ \(2x+3=5(x-1)=5x-5\)

→ \(8=3x\) → \(x=\dfrac{8}{3}\)

✓ \(f^{-1}(5) = \dfrac{8}{3}\)

💡 Solution

Set \(f(x)=5\): \(2x+3=5x-5\) → \(3x=8\) → \(f^{-1}(5)=\dfrac{8}{3}\)

G-01

Similar Triangles Easy

Two similar triangles have sides 6, 8, 10 and 9, ?, 15.
Find the missing side.
Enter a single integer.

⚡

Memory Key SIMILAR → find scale factor (ratio), then multiply all sides by it

Steps

Scale factor: \(9 \div 6 = 1.5\) (or \(15 \div 10 = 1.5\) ✓ consistent)

→ Missing side: \(8 \times 1.5 = 12\)

💡 Solution

Scale factor = 9/6 = 3/2. Missing side = 8 × 3/2 = 12

G-02

Circle Theorems Medium

An inscribed angle in a circle intercepts an arc of 130°.
What is the measure of the inscribed angle?
Enter degrees, number only.

⚡

Memory Key INSCRIBED ANGLE = ½ × (intercepted arc) — always half the arc!

Formula Reminder

Inscribed angle \(= \dfrac{1}{2}\) × intercepted arc

→ \(\dfrac{1}{2} \times 130° = 65°\)

💡 Solution

Inscribed angle = ½ × 130° = 65°

G-03

Right Triangles & Trigonometry Medium

In right triangle ABC with right angle at C,
\(\sin A = \dfrac{5}{13}\). Find \(\cos A\).
Enter as fraction.

⚡

Memory Key PYTHAGOREAN IDENTITY → sin²A + cos²A = 1 → cos²A = 1 − sin²A

Steps

\(\sin A = 5/13\) → opposite=5, hypotenuse=13

→ adjacent = \(\sqrt{13^2-5^2}=\sqrt{144}=12\)

✓ \(\cos A = 12/13\)

💡 Solution

Hyp=13, opp=5 → adj=12. cos A = 12/13

G-04

Volume & Surface Area Easy

A cylinder has radius 3 and height 7.
Find its lateral surface area.
Leave answer in terms of π. Enter: n π (e.g. "12pi")

⚡

Memory Key LATERAL = 2πrh (think: "unroll" the cylinder into a rectangle → width=circumference)

Formula

Lateral S.A. \(= 2\pi r h\)

→ \(= 2\pi(3)(7) = 42\pi\)

💡 Solution

\(2\pi(3)(7) =\) 42π

G-05

Coordinate Geometry Medium

Find the equation of a circle with center (−2, 3) and passing through (1, 7).
What is the radius? (Enter a number, exact form OK e.g. "5")
Enter the radius value.

⚡

Memory Key CIRCLE → (x−h)²+(y−k)²=r² | radius = distance from center to point on circle

Steps

\(r = \sqrt{(1-(-2))^2 + (7-3)^2}\)

→ \(= \sqrt{9+16} = \sqrt{25} = 5\)

💡 Solution

\(r=\sqrt{9+16}=\sqrt{25}=\)5

G-06

Parallel Lines & Angles Easy

Two parallel lines are cut by a transversal.
One co-interior (same-side interior) angle is \( (3x+10)° \)
and the other is \( (2x+20)° \).
Find \( x \).
Enter a single integer.

⚡

Memory Key CO-INTERIOR (same side) = SUPPLEMENTARY → they add up to 180°

Steps

\((3x+10)+(2x+20) = 180\)

→ \(5x+30=180\) → \(5x=150\) → \(x=30\)

💡 Solution

Co-interior angles sum to 180°: \(5x+30=180\) → x = 30

G-07

Triangle Centers Hard

The centroid of a triangle divides each median in the ratio
2:1 from vertex to midpoint.
If a median has total length 18, how far is the centroid from the vertex?
Enter a single integer.

⚡

Memory Key CENTROID = "2/3 from vertex" → always 2/3 of the median length from vertex end

Explanation

Centroid divides median → vertex part : midpoint part = 2:1

→ Vertex distance = \(\dfrac{2}{3} \times 18 = 12\)

💡 Solution

\(\dfrac{2}{3} \times 18 =\) 12

G-08

Transformations Medium

Point \( P = (3, -2) \) is rotated 90° counterclockwise about the origin.
What are the new coordinates?
Enter as (x, y)

⚡

Memory Key 90° CCW → (x, y) becomes (−y, x) | 90° CW → (x, y) becomes (y, −x)

Rotation Rules Cheat Sheet

90° CCW: \((x,y) \to (-y, x)\)

180°: \((x,y) \to (-x,-y)\)

90° CW: \((x,y) \to (y,-x)\)

→ \((3,-2)\) 90° CCW → \((2, 3)\)

💡 Solution

Rule: (x,y) → (−y, x). So (3,−2) → (2, 3)

G-09

Trigonometric Ratios (Law of Sines) Medium

In triangle ABC: \( a = 8,\; \angle A = 30°,\; \angle B = 45° \).
Use the Law of Sines to find side \( b \).
Enter exact simplified form: e.g. "4sqrt(2)"

⚡

Memory Key LAW OF SINES → a/sin A = b/sin B → cross multiply → solve for b

Steps

\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)

→ \(\dfrac{8}{\sin 30°} = \dfrac{b}{\sin 45°}\)

→ \(\dfrac{8}{0.5} = \dfrac{b}{\frac{\sqrt{2}}{2}}\) → \(16 = \dfrac{2b}{\sqrt{2}}\)

✓ \(b = 8\sqrt{2}\)

💡 Solution

\(b = 8 \cdot \dfrac{\sin 45°}{\sin 30°} = 8 \cdot \dfrac{\sqrt{2}/2}{1/2} = 8\sqrt{2}\)
b = 8√2

G-10

Geometric Proofs / Quadrilaterals Hard

In parallelogram ABCD, \(\angle A = (4x+10)°\) and \(\angle B = (2x+30)°\).
Find \(\angle A\) in degrees.
Enter degrees, number only.

⚡

Memory Key PARALLELOGRAM → consecutive angles are SUPPLEMENTARY (add to 180°)

Steps

Consecutive angles in a parallelogram sum to 180°:

\((4x+10)+(2x+30) = 180\)

→ \(6x+40=180\) → \(6x=140\) → \(x=\dfrac{70}{3}\)

→ \(\angle A = 4(\frac{70}{3})+10 = \frac{280}{3}+10 = \frac{310}{3} \approx 103.3°\)

💡 Solution

\(6x=140 \Rightarrow x=70/3\)
\(\angle A = 4(70/3)+10 = 280/3+30/3 =\) 310/3 ≈ 103.3°